# Print Cube Root of A Number in C++

Confronting with numbers is the first thing each one of us has learned in school. However, as we step up high more and more difficult questions based on the same fundamentals seem to appear in our way. It’s the same with programming, using the same fundamentals to find solutions for multiple complex problems. Now the cool thing is, you can be creative in your way eventually coming up with a unique solution. And if you’re able to do this, then you’ve clearly understood how to use programming for others and your good. But keeping that aside, we’ve talked about numbers before. So what if we find the cube root of numbers, seems too easy right!. Well, So let’s quickly find out how to Print the Cube Root of A Number in C++.

With the help of this program, you will be able to find the root square of integers. Practising these types of questions also helps to get an upper edge in Competitive Programming.

## What’s The Approach?

- We’ll use
**Binary Search Algorithm**to find the cube root of the given input number**n.**

- Therefore we’ll initialize

&**start=0**

, and we’ll find the middle value in this list by executing**end=n****(start + end)/2**

- We’ll
**set an error point e**, below which if the calculation goes will be**set as the cube root**of n. So the condition is**(n – mid*mid*mid) < e**

- However, if

we’ll set**(mid*mid*mid)>n**

. And if**end=mid**

we’ll set**(mid*mid*mid)<n****start=mid**

**Also Read: Implement Linear Search using C++**

## C++ Program To Print Cube Root of A Number

**Input:**

**n = 8**

**Output:**

**The cubic root of 8.000000 is 2.000000**

// C++ program to find cubic root of a number #include <bits/stdc++.h> using namespace std; // Returns the absolute value of n-mid*mid*mid double diff(double n,double mid) { if (n > (mid*mid*mid)) return (n-(mid*mid*mid)); else return ((mid*mid*mid) - n); } // Returns cube root of a no n double cubicRoot(double n) { // Set start and end for binary search double start = 0, end = n; // Set precision double e = 0.0000001; while (true) { double mid = (start + end)/2; double error = diff(n, mid); // If error is less than e then mid is // our answer so return mid if (error <= e) return mid; // If mid*mid*mid is greater than n set // end = mid if ((mid*mid*mid) > n) end = mid; // If mid*mid*mid is less than n set // start = mid else start = mid; } } // Driver code int main() { double n = 8; printf("Cubic root of %lf is %lf\n", n, cubicRoot(n)); return 0; }